Answer :
To identify the spectator ions in the given chemical equation, we need to examine the equation and see which ions do not undergo any change from the reactant side to the product side. Spectator ions are essentially ions that appear unchanged on both sides of a chemical equation.
Given chemical equation:
[tex]\[ 2 H^{+} + SO_{4}^{2-} + Ca^{2+} + 2 I \rightarrow CaSO_{4} + 2 H^{+} + 2 I \][/tex]
Let's analyze this step-by-step:
1. Reactants:
- \(2 H^{+}\)
- \( SO_{4}^{2-}\)
- \( Ca^{2+}\)
- \( 2 I\) (it was mentioned that "r" might be a typographical error, so we'll consider it as \( I \), iodine ion.)
2. Products:
- \( CaSO_{4} \)
- \( 2 H^{+} \)
- \( 2 I\)
3. Comparison:
- \( 2 H^{+} \) ions are present on both the reactant side and the product side.
- \( 2 I\) ions are also present on both sides without any change.
Therefore, the ions that are unchanged and present on both the reactant and product sides are \( H^{+} \) and \( I \).
These ions do not participate in the actual reaction and hence are considered spectator ions.
So, the correct answer is:
A. [tex]\( H^{+} \)[/tex] and [tex]\( I \)[/tex]
Given chemical equation:
[tex]\[ 2 H^{+} + SO_{4}^{2-} + Ca^{2+} + 2 I \rightarrow CaSO_{4} + 2 H^{+} + 2 I \][/tex]
Let's analyze this step-by-step:
1. Reactants:
- \(2 H^{+}\)
- \( SO_{4}^{2-}\)
- \( Ca^{2+}\)
- \( 2 I\) (it was mentioned that "r" might be a typographical error, so we'll consider it as \( I \), iodine ion.)
2. Products:
- \( CaSO_{4} \)
- \( 2 H^{+} \)
- \( 2 I\)
3. Comparison:
- \( 2 H^{+} \) ions are present on both the reactant side and the product side.
- \( 2 I\) ions are also present on both sides without any change.
Therefore, the ions that are unchanged and present on both the reactant and product sides are \( H^{+} \) and \( I \).
These ions do not participate in the actual reaction and hence are considered spectator ions.
So, the correct answer is:
A. [tex]\( H^{+} \)[/tex] and [tex]\( I \)[/tex]
